# Linear Forces Speed Review

###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

###### Article Content
In this section, we will master Newton’s Second Law (F = ma). We will also cover common areas of struggle, such as drawing FBDs and finding net forces.

### Concept overview

1. Newton’s force laws
2. Drawing FBDs properly
3. Resolving vectors
4. Finding the net force
5. Solving problems

### Newton’s Laws

We can use Newton’s three laws to help explain everyday phenomena, such as:

• What force, if any, is needed to make a tennis ball bounce off the sidewalk?
• If you walk along a log floating on a lake, why does the log move in the opposite direction
• Why do you seem to be thrown forward in a car that rapidly decelerates?

#### Law 1: Inertia

An object in motion will stay in motion. An object at rest will stay at rest.

Alternate explanation: all objects want to resist changes in their motion. The more inertia (mass) an object has, the more it can resist change. Think of how hard it is to bring a plane to a complete stop.

#### Law 2: F_{net} = ma

Any horizontal or vertical NET force acting on a mass will cause an acceleration.

Vertical net force is the sum of ALL forces vertically. The horizontal net force is the sum of ALL forces horizontally.

#### Law 3: Equal and Opposite Forces

Push down on your desk. Does it move? It doesn’t, because it’s pushing back up at you with the same force. When two objects interact with each other, they both apply the same force to each other.

For every action, there is an equal and opposite reaction.

Alternate explanation: think of a truck hitting a motorcycle. They will experience the same force, but since the motorcycle has less mass it will experience greater acceleration.

### FBD

If you want to solve questions correctly, your best bet is to start by drawing an FBD (force body diagrams). Here are the rules for drawing common forces.

1. Weight ALWAYS points straight down. Draw the vector from the center of the object.
2. The normal force is ALWAYS 90° to the surface. Draw the vector from the bottom of the object.
3. Friction always points opposite to the direction of motion (except in circular motion, in which it points inwards)
4. Vectors can be at angles, but don’t draw the vector components unless asked to.
5. Vectors of equal magnitude should be of equal length.

#### Common FBDs

Know how to draw the following FBDs:

• Mass in free fall, neglecting air resistance
• Mass falling at terminal velocity
• Box sliding horizontally with friction
• Box sliding down a ramp with friction
• Elevator going up and slowing down
• Two different masses on opposite sides of a pulley

### Mass vs. Weight

Mass is a scalar quantity (measured in kg). It remains constant everywhere in the universe.

Weight is a vector quantity (measured in Newtons). It is the force on a mass due to gravity. Weight changes with gravity.

\text{Weight} = F_{\text{net}} = mg

### Setting Directions

If you’ve finished Unit 1 Kinematics, you’ll be familiar with setting direction.

Gravity can be positive or negative. The direction of positive and negative motion is arbitrarily chosen. That means you can choose if up means +/- or if left means +/-.

But you need to be consistent. Indicate the directions on your FBD. If you set down to be positive, then gravity will be positive since it points down.

### Resolving Vectors

We only use horizontal and vertical forces when solving force problems. Therefore, if you are using a vector that is at an angle, you need to break (resolve) it into x and y components.

Use this trig shortcut: Given the vector \vec{A} , look at the component of the triangle you are trying to find.

• If that component is opposite to the angle → use \vec{A} \sin \theta
• If that component is adjacent to the angle → use \vec{A} \cos \theta

If you’re still confused on vector components, read this short guide.

You can also watch the video below.

### Net Forces vs. Equilibrium

The net force is the SUM of ALL forces in either the X or Y direction.

An object in equilibrium experiences zero net force in all directions. In other words, any force acting in a particular direction is canceled out with an equal but opposite force along the same axis.

Since F_{net} = 0 = ma → this implies the net acceleration of the object is 0.

This either means the object is not moving OR moving at constant velocity. For example, a book laying on the table has 0 net force, because the weight vector and normal vector cancel out.

→ 0 Net Force = 0 acceleration = constant velocity

### Common Equations

Kinetic Friction Force: f_k = \mu_kN

Static Friction Force: f_s = \mu_sN

• N = normal force
• \mu_k = coefficient of kinetic friction (for when the object is moving)
• \mu_s = coefficient of static friction (for when the object is not moving
• \mu_s > \mu_k
• PRO TIP: on a ramp –> \mu_s = \tan\theta

Hooke’s Law (Spring Force): F_s = kx

• k = spring constant
• x = compression or stretch of the spring

Centripetal force: F_c = ma_c = \frac{mv^2}{r}

• v = tangential velocity
• r = radius of the circle

Gravitational Force between two masses: F_g = \frac{Gm_1m_2}{r^2}

• G = gravitational constant ≈ 6.67 × 10-11
• m_1 = mass of object 1
• m_2 = mass of object 2
• r = distance between m_1 and m_2

### Solving Problems Quickly

Students struggle the most on this. Using the steps (framework) below you can solve ALL force problems, it just comes down to practice. This framework was covered in depth in the Unit 2.4 course article.

1. Turn the word problem into a FBD.
2. Use the FBD to find either the horizontal or vertical net force.
3. Lastly set the net force equal to ma, to solve for the unknown variable.

There are many of types of problems involving forces. But I’ve simplified it to this list of 60 linear force questions [available for course members], based on the actual AP Physics 1 exam. These will likely show up on your test.

#### Common Types of Linear Force Questions

Listed below is a list of common linear force problems (in the next article we’ll add circular force problems). You should know how to solve all of these, using the 3 step method above.

As a bonus, I’ve linked short guides on how to specifically tackle each type of problem.

### Extra Help

If any of this sounds confusing, you’re not alone! I’ve helped 100s of Physics students to achieve a 5 and boost class grades. You’ll start seeing results in just 3 lessons or less, guaranteed.

Book A Free 1-to-1 Trial Lesson

### 10 Practice Questions for Mastery

Question 1
Solve Type - Mathematical
Find the downwards acceleration of an elevator, given that the ratio of a person’s stationary weight to their weight in the elevator is 5:4.
Question 2
Difficulty - Intermediate
Solve Type - Mathematical
Determine the force needed to push a 150 kg body up a smooth 30° incline with an acceleration of 6 m/s2.
Question 3
Difficulty - Intermediate
Solve Type - Mathematical
A 1kg and unknown mass (M) hangs on opposite sides of the pulley suspended from the ceiling. When the masses are released, M accelerates down at 5 m/s². What is M?
Question 4
Difficulty - Intermediate
Solve Type - Mathematical
A sled moves with constant speed down a sloped hill. The angle of the hill with respect to the horizontal is 10.0°. What is the coefficient of kinetic friction between the sled and the hill’s surface?
Question 5
Difficulty - Intermediate
Solve Type - Conceptual
A ladder is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on the ladder by the floor?
1. The force of gravity on the ladder.

2. The normal force exerted on the ladder by the floor.

3. The frictional force exerted on the ladder by the wall.

4. The normal force exerted on the ladder by the wall.

5. None of these choices.

Question 6
Solve Type - Mathematical
Three blocks of masses 5, 4, and 3 kg are placed side by side in that order. A 25 N force applied on the 5 kg block accelerates all three blocks together to the right. Find the acceleration of the blocks and the normal force the 4 kg block exerts on the 3 kg block.
Question 7
Difficulty - Intermediate
Solve Type - Conceptual
A truck of mass 3500 kg hits the back of a small car of mass 1400 kg. Which car exerted more force on the other and why?
1. The truck exerts more force on the car because it has more mass.

2. The tuck exerts more force on the car because it has more acceleration.

3. The car exerts more force on the truck because it has less mass.

4. The car exerts more force on the truck because it has more acceleration

5. They exert the same force, and the resulting acceleration of the car is greater.

Question 8
Difficulty - Intermediate
Solve Type - Mathematical
A block of mass m is accelerated across a rough surface by a force of magnitude F exerted at an angle θ above the horizontal. The frictional force between the block and surface is ƒ. Find the acceleration of the block (as an equation).
Question 9
Difficulty - Intermediate
Solve Type - Conceptual
A clothesline is stretched between two trees. A tire hangs in the middle of the line, and the two halves of the line make equal angles with the horizontal. The tension in the line is
1. half the tire’s weight.

2. is equal to the tire’s weight divided by 9.18 m/s2.

3. is less than half the tire’s weight.

4. is equal to the tire’s weight.

5. is more than the tire’s weight.

Question 10
Difficulty - Intermediate
Solve Type - Proportional Analysis
A child on Earth has a weight of 500N. Determine the weight of the child if the earth was to triple in both mass and radius (3M and 3r).
1. 2 m/s2
2. 1650 N
3. 3 kg
4. tan(10) or .176
5. C.
6. 2.1 m/s2 and 6.25 N
7. E.
8. a = (Fcosθ – ƒ)/m
9. E.
10. 167 N

### Practice Exam

If you’re feeling confident try completing this practice exam. Make sure to time yourself and check your answers in the end.

Here’s another Multiple Choice Test, that has questions similar to the AP Exam.

### Next Speed Review

In the next speed review we’ll cover forces that cause a circular motion. We often call this centripetal forces.

###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

## Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy

Unit 5 – Momentum

Unit 6 – Torque

Unit 7 – Oscillations

Unit 8 – Fluids

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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